Question

A cylindrical vessel is filled upto a height h with a liquid of density p. The cross sectional area of the vessel is
a. From a small orifice near the bottom if some liquid flows out so that the level falls to h/2the wotk done by the gravity in doing so

Answer 1

Answer:

Work done by gravity to make the level of water to be half is given as

$W = \frac{3\rho a h^2 g}{8}$

Explanation:

Initial total mass of the liquid in the container

$m = \rho V$

$m_1 = \rho (a)(h)$

now when small orifice is open at the bottom then final level of liquid becomes half

So final mass of the liquid

$m_2 = \rho V'$

$m_2 = \rho a \frac{h}{2}$

now change in the potential energy of the liquid is total work done by the gravity

So here we have

$W = m_1 g\frac{h}{2} - m_2 g \frac{h}{4}$

$W = \frac{\rho a h^2g}{2} - \frac{\rho a h^2 g}{8}$

$W = \frac{3\rho a h^2 g}{8}$

#Learn

Topic : Work energy relation

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